What is the time of flight for a projectile launched from a cliff?

[brandon26]

A stone is projected from a point O on a cliff with a speed of of 20ms^-1 at an angle of elevation of 30. T seconds later the angle of depression of the stone from O is 45. Find the value of T.

I assumed that when the stone is at depression of 45 degree to O, it has reached the horizontal plane again, hence displacement is 0. so I used s = ut + 0.5at^2.
s=0
a=-9.8
u=20sin30
therefore t= 2.04, which is apparabtly wrong.

Please help.

 

[Päällikkö]

You've assumption of the angle is wrong.
$$\tan \theta = \frac{v_y}{v_x}$$

 

[brandon26]

You've assumption of the angle is wrong.
$$\tan \theta = \frac{v_y}{v_x}$$

Could you expand on that please?

 

[Päällikkö]

http://img29.imageshack.us/my.php?image=img0071wo.jpg
That's the situation at some point during the flight.

 

[brandon26]

But how can I show that angle of depression is 45 degrees on that diagram?

 

[Päällikkö]

You'll want theta = -45^o (depending on the positive/negative directions you've chosen).

On the other hand, as tan -45^o = -1, v_x = -v_y (at the desired instant).

 

[brandon26]

Sorry I am confused! Could you go through the question please?

 

[Päällikkö]

Positive/negative directions:
+y
|
|
----+x

$$\tan \theta = \frac{v_y}{v_x}$$
We also know that
$$v = v_0+at$$


$$\tan \theta = \frac{v_{0y}-gt}{v_{0x}}$$